$\overline{AC}$ is $6$ units long $\overline{BC}$ is $8$ units long $\overline{AB}$ is $10$ units long What is $\cot(\angle BAC)?$ $A$ $C$ $B$ $6$ $8$ $10$
Solution: $\cot(\angle BAC) = \dfrac{1}{\tan(\angle BAC)}$ How can we find $\tan(\angle BAC)$ SOH CAH TOA angent = pposite over djacent Opposite $= \overline{BC} = 8$ Adjacent $= \overline{AC} = 6$ $\tan(\angle BAC) = \dfrac{8}{6}$ $\cot(\angle BAC) = \dfrac{1}{\tan(\angle BAC)} = \dfrac{6}{8}$